Sixty-three knights on a chessboard

Question

_{This puzzle is part of the Monthly Topic Challenge #7: Board games.}

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If we're using rooks, then this would be equivalent to a classic "fifteen puzzle" though on a larger grid. This time we'll use knights instead of rooks.
Place sixty-three knights, each in a different color, on the 8x8 chessboard except for one of the corner squares.
Is it possible to use a finite number of moves to achieve the goal of simply swapping two of the knights on the grid with the other knights back where they started?
If yes, find the smallest number of moves to achieve the goal. You do not need to alternate players/colors.

Note: Capturing is not allowed here.

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Solve the task described above.

(optional) What if we're using queens or bishops?

Answer 1

It's

not possible for pretty much the same reason as with the classic 15 puzzle. Treating the gap as a "special colour" piece each move amounts to swapping two pieces. The parity of the permutation of pieces resulting from making several moves therefore alternates between odd and even. At the same time the colour of the square the gap (or special colour piece) sits on also alternates. So, the gap will be on a black square (if we start with the top right corner empty) whenever the permutation is even and on a white square whenever it is odd. But swapping two (non-gap) pieces would be an odd permutation with the gap still on its black origin. So it is not possible

Optional bits:

With bishops

it's also not possible, same argument, only we need a custom colouring of the board.

With queens

it is possible, for example we can show that we construct every swap of adjacent queens or some other set of generators of the full permutation group.